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Svetradugi [14.3K]
3 years ago
10

(1-\sqrt{x})/1+\sqrt{x}

Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

\frac{1 - 2 \sqrt{x} + x }{1 - x}

Step-by-step explanation:

\frac{1 -  \sqrt{x} }{1 +  \sqrt{x} }

Rationalize the denominator by multiplying by it conjugate.

\frac{1 -  \sqrt{x} }{1 +  \sqrt{x} }  \times  \frac{1 -  \sqrt{x} }{1 -  \sqrt{x} }  =  \frac{1 - 2 \sqrt{x}  + x}{1 - x}

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Line u has a slope of<br> -2/5 is parallel to line u. What is the slope of line v?
tresset_1 [31]

Answer:

-2/5

Step-by-step explanation:

The slope will be same , that is -2/5 .

7 0
3 years ago
joe earns R25 per hour working at a pizza shop. How much will he earn over a weekend of two shifts, each six hours long​
NeX [460]

Answer:

R300

Step-by-step explanation:

<u>To solve:</u>

  • Multiply 25 by 6 to represent the earnings of a shift.
  • Multiply the shift earnings by 2 to represent how much he earned over the weekend.

<u>Multiply</u><u> </u><u>25 by 6:</u>

25 \times 6 = 150

<u>Multiply 150 by 2:</u>

<u>150 \times 2 = 300</u>

Joe will earn R300 over the weekend,

6 0
3 years ago
Jack plans on using network ATMs about 4 times per month. What would be Jack’s total estimated annual fees for a checking accoun
anyanavicka [17]

Answer:

$104

Step-by-step explanation:

Thank me...

3 0
4 years ago
Read 2 more answers
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
( - 7a'y + xy + 3x + 2) + (5r²y – 5cy - 6x – 7)
AlladinOne [14]

Answer:

Adding ( - 7a'y + xy + 3x + 2) + (5r²y – 5cy - 6x – 7) will equal to

−7ya'+xy+5r2y−5cy−3x−5

8 0
3 years ago
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